Question

Please solve this?

A water tank is formed by the combination of cylinder and hemisphere. The total height of the tank is 20m and base area is 154 sq. m. If the tank is filled with water at the rate of 45 paisa per litre, what is the total cost to fill the water?

Answer 1

`approx 122426730 text{ P}`

Explanation:

Not totally sure what's intended here. The hemisphere's volume is `1/2 (4/3 pi r^3) = 2/3 pi r^3` and the cylinder's volume is `pir^2 h=pi r^2(20-r)= 20 pi r^2 - pi r^3` so a total volume of

`V = 20 pi r^2 - pi/3 r^3`

Not sure what a base area of 154 sq meters means, let's assume it means

`154 = pi r ^2`

`r^2=154/pi`

`r = sqrt{154/pi}`

`V = 20 pi (154/pi) - pi/3 (154/pi) sqrt{154/pi}`

`V = 154/3 (60 - sqrt(154/π) ) approx 2720.594 text{m}^3`

`text{cost} approx 45 text{P}/text{L} times 1000 text{L}/text{m}^3 times 2720.594 text{m}^3 approx 122,426,730 text{ P}`

Answer 2

I'm assuming we're dealing in Rupees here, which means that the total cost is 1,224,300.00 Rupees (122,430,000 paise)

Explanation:

The first thing to do is figure out an expression to determine the volume of the water tank.

I'm assuming that the tank is an upright cylinder that is capped by a hemisphere. The total volume can be expressed as:

`V_"tot"=V_"cyl"+V_"hemi"`

The volume of a Cylinder is `hpir^2`, where `h` is the height of the cylinder.

The volume of a hemisphere is half of the volume of a sphere:

`V_"sphere"=4/3pir^3`

`V_"hemi"=V_"sphere"/2=(4/3pir^3)/2`

`V_"hemi"=2/3pir^3`

`V_"tot"=hpir^2+2/3pir^3`

We also know that h is the height of the tank MINUS the radius of the hemisphere, since it is capped as one:

`h=20-r`

`V_"tot"=(20-r)pir^2+2/3pir^3`

Rearranging:

`V_"tot"=pir^2(20-r+2/3r)`

`V_"tot"=pir^2(20-r/3)`

We also know that the base area is the area of the circle of the cylidner, which is equivalent to `pir^2`

`V_"tot"=154(20-r/3)`

Let's solve for `r` to calculate the total volume:

`154=pir^2 rArr r=sqrt(154/pi)`

`r~=7`

`(r=7.001409)`

Now that we know `r`:

`V_"tot"=154(20-7/3)`

`V_"tot"=154(53/3)`

`color(blue)(V_"tot"=2720 2/3 m^3`

now that we know the volume in cubic meters, we need to convert to liters, to match units for the cost per liter:

`1m^3=1000L`

`color(blue)(2720 2/3m^3)=2720666 2/3L`

Finally, we have the cost per liter, which will give us our final cost:

`"COST"=45"paisa"/cancel(L)xx2720666 2/3cancel(L)`

`"COST"=122430000" paisa"`

Assuming we are dealing in Rupees, 1 Rupee is 100 paise:

`"COST"=1224300" Rupee"`